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Finding the general solution to a differential equation

  1. Dec 8, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\frac{d^{2}y}{dt}[/tex] +4[tex]\frac{dy}{dt}[/tex]+20y=e[tex]^{-2t}[/tex](sin4t+cos4t)

    2. Relevant equations



    3. The attempt at a solution

    The solution to the homogeneous equation: [tex]\frac{d^{2}y}{dt}[/tex] +4[tex]\frac{dy}{dt}[/tex]+20y=0 is

    y= k1e[tex]^{-2t}[/tex]cos4t +k2e[tex]^{-2t}[/tex]sin4t

    Then I guessed ae[tex]^{-2+4i}[/tex] as a possible solution and it didn't work, and that's where I'm stuck.
     
  2. jcsd
  3. Dec 8, 2009 #2

    LCKurtz

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    Since the complementary solution yc= e-2t(Acos(4t) + Bsin(4t)) is repeated in the non-homogeneous term, for your particular solution try

    yp = te-2t(Ccos(4t) + Dsin(4t))
     
  4. Dec 9, 2009 #3
    Nope...didn't work. :(
     
  5. Dec 9, 2009 #4

    LCKurtz

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    Yes, it does work. Check your work or show it here if you can't find your error.
     
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